Solve Applications of Quadratic Equations

By the end of this section, you will be able to: solve applications modeled by quadratic equations.

Solve Applications Modeled by Quadratic Equations

We solved some applications that are modeled by quadratic equations earlier, when the only method we had to solve them was factoring. Now that we have more methods to solve quadratic equations, we will take another look at applications.

Let’s first summarize the methods we now have to solve quadratic equations.

Methods to Solve Quadratic Equations

Factoring
Square Root Property
Completing the Square
Quadratic Formula

As you solve each equation, choose the method that is most convenient for you to work the problem. As a reminder, we will copy our usual Problem-Solving Strategy here so we can follow the steps.

Use a Problem-Solving Strategy.

Read the problem. Make sure all the words and ideas are understood.
Identify what we are looking for.
Name what we are looking for. Choose a variable to represent that quantity.
Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
Solve the equation using algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

We have solved number applications that involved consecutive even and odd integers, by modeling the situation with linear equations. Remember, we noticed each even integer is 2 more than the number preceding it. If we call the first one $n$ , then the next one is $n+2$ . The next one would be $n+2+2$ or $n+4$ . This is also true when we use odd integers.

Consecutive even integers: 64, 66, 68	Consecutive odd integers: 77, 79, 81
$n$ : first even integer	$n$ : first odd integer
$n+2$ : second consecutive even integer	$n+2$ : second consecutive odd integer
$n+4$ : third consecutive even integer	$n+4$ : third consecutive odd integer

Some applications of odd or even consecutive integers are modeled by quadratic equations. The notation above will be helpful as you name the variables.

Example 9.35

The product of two consecutive odd integers is 195. Find the integers.

Solution.

Step 1. Read the problem.

Step 2. Identify what we are looking for. We are looking for two consecutive odd integers.

Step 3. Name what we are looking for. Let $n$ be the first odd integer and $n+2$ the next odd integer.

Step 4. Translate into an equation. “The product of two consecutive odd integers is 195.”

n(n+2)=195

Step 5. Solve the equation.

\begin{array}{lrcl} \text{Distribute.} & n^2+2n &=& 195 \\[4pt] \text{Write in standard form.} & n^2+2n-195 &=& 0 \\[4pt] \text{Factor.} & (n+15)(n-13) &=& 0 \\[4pt] \text{Use the Zero Product Property.} & n+15=0 &\text{ or }& n-13=0 \\[4pt] \text{Solve each equation.} & n=-15 &\text{ or }& n=13 \end{array}

There are two values of $n$ that are solutions. This will give us two pairs of consecutive odd integers for our solution: $13,15$ and $-15,-13$ .

Step 6. Check the answer. Both pairs are consecutive odd integers, and $13\cdot15=195$ and $(-13)(-15)=195$ .

Step 7. Answer the question. Two consecutive odd integers whose product is 195 are 13, 15 and $-13,-15$ .

The product of two consecutive odd integers is 99. Find the integers. Enter the positive pair followed by the negative pair, separated by commas.

The product of two consecutive even integers is 168. Find the integers. Enter the positive pair followed by the negative pair, separated by commas.

Area of a Triangle. For a triangle with base, $b$ , and height, $h$ , the area, $A$ , is given by the formula

A=\tfrac12bh.

Recall that when we solve geometric applications, it is helpful to draw the figure.

Example 9.36

An architect is designing the entryway of a restaurant. She wants to put a triangular window above the doorway. Due to energy restrictions, the window can only have an area of 120 square feet and the architect wants the base to be 4 feet more than twice the height. Find the base and height of the window.

Solution.

Step 1. Read the problem. Draw a picture: the triangular window has height $h$ and base $2h+4$ .

Step 2. Identify what we are looking for. We are looking for the base and height.

Step 3. Name what we are looking for. Let $h$ be the height of the triangle. Then $2h+4$ is the base.

Step 4. Translate into an equation. Use $A=\tfrac12bh$ .

Step 5. Solve the equation.

\begin{array}{lrcl} \text{Substitute the values.} & 120 &=& \tfrac12(2h+4)h \\[10pt] \text{Distribute.} & 120 &=& h^2+2h \\[4pt] \text{Write in standard form.} & h^2+2h-120 &=& 0 \\[4pt] \text{Factor.} & (h-10)(h+12) &=& 0 \\[4pt] \text{Use the Zero Product Property.} & h-10=0 &\text{ or }& h+12=0 \\[4pt] \text{Simplify.} & h=10 &\text{ or }& h=-12 \end{array}

Since $h$ is the height of a window, $h=-12$ does not make sense. The height is 10 feet, and the base is $2(10)+4=24$ feet.

Step 6. Check the answer. A triangle with height 10 and base 24 has area $\tfrac12(24)(10)=120$ . Yes.

Step 7. Answer the question. The height of the triangular window is 10 feet and the base is 24 feet.

Find the base and height of a triangle whose base is four inches more than six times its height and has an area of 456 square inches. Enter the base and height, separated by a comma.

A triangle has an area of 110 square feet and a base that is two feet less than twice the height. Enter the base and height, separated by a comma.

In the two preceding examples, the number in the radical in the Quadratic Formula was a perfect square and so the solutions were rational numbers. If we get an irrational number as a solution to an application problem, we will use a calculator to get an approximate value.

We will use the formula for the area of a rectangle to solve the next example.

Area of a Rectangle. For a rectangle with length,

L

, and width,

W

, the area,

A

, is given by the formula

A=LW

Example 9.37

Mike wants to put 150 square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than 3 times the width. Find the length and width. Round to the nearest tenth of a foot.

Solution.

Step 1. Read the problem. Draw a rectangle with width $w$ and length $3w-1$ .

Step 2. Identify what we are looking for. We are looking for the length and width.

Step 3. Name what we are looking for. Let $w$ be the width. Then $3w-1$ is the length.

Step 4. Translate into an equation. Use $A=L\cdot W$ .

Step 5. Solve the equation.

\begin{array}{lrcl} \text{Substitute the values.} & 150 &=& (3w-1)w \\[4pt] \text{Distribute.} & 150 &=& 3w^2-w \\[4pt] \text{Write in standard form.} & 3w^2-w-150 &=& 0 \end{array}

Here $a=3$ , $b=-1$ , and $c=-150$ . Using the Quadratic Formula,

\begin{aligned} w&=\frac{-(-1)\pm\sqrt{(-1)^2-4(3)(-150)}}{2(3)}\\ &=\frac{1\pm\sqrt{1801}}6. \end{aligned}

The two solutions are approximately $7.2$ and $-6.9$ . We eliminate the negative solution for the width. The width is approximately 7.2 feet, and the length is $3(7.2)-1\approx20.6$ feet.

Step 6. Check the answer. Since the answers are approximate, the area will not come out exactly to 150.

Step 7. Answer the question. The width is approximately 7.2 feet and the length is approximately 20.6 feet.

The length of a 200-square-foot rectangular vegetable garden is four feet less than twice the width. Find the length and width, to the nearest tenth of a foot. Enter the length and width, separated by a comma.

A rectangular tablecloth has an area of 80 square feet. The width is 5 feet shorter than the length. Find the length and width to the nearest tenth of a foot. Enter the length and width, separated by a comma.

The Pythagorean Theorem gives the relation between the legs and hypotenuse of a right triangle. We will use the Pythagorean Theorem to solve the next example.

Pythagorean Theorem. In any right triangle, where

a

and

b

are the lengths of the legs, and

c

is the length of the hypotenuse,

a^2+b^2=c^2

Example 9.38

Rene is setting up a holiday light display. He wants to make a “tree” in the shape of two right triangles and has two 10-foot strings of lights to use for the sides. He will attach the lights to the top of a pole and to two stakes on the ground. He wants the height of the pole to be the same as the distance from the base of the pole to each stake. How tall should the pole be?

Solution.

Step 1. Read the problem. Draw a picture of the two right triangles.

Step 2. Identify what we are looking for. We are looking for the height of the pole.

Step 3. Name what we are looking for. Let $x$ be both the height of the pole and the distance from the pole to a stake. Each side is a right triangle with legs $x,x$ and hypotenuse 10.

Step 4. Translate into an equation. Use the Pythagorean Theorem.

Step 5. Solve the equation.

\begin{array}{lrcl} \text{Substitute.} & x^2+x^2 &=& 10^2 \\[4pt] \text{Simplify.} & 2x^2 &=& 100 \\[4pt] \text{Divide by 2.} & x^2 &=& 50 \\[4pt] \text{Use the Square Root Property.} & x &=& \pm\sqrt{50} \\[4pt] \text{Simplify the radical.} & x &=& \pm5\sqrt2 \end{array}

We eliminate the negative solution. If we approximate to the nearest tenth, $x\approx7.1$ .

Step 6. Check the answer in the Pythagorean Theorem.

Step 7. Answer the question. The pole should be about 7.1 feet tall.

A flag pole is three times the length of its shadow. The distance between the end of the shadow and the top of the pole is 20 feet. Find the shadow length and pole length to the nearest tenth. Enter them in that order, separated by a comma.

The distance between opposite corners of a rectangular field is four more than the width. The length is twice the width. Find the distance between opposite corners, to the nearest tenth.

The height of a projectile shot upward from the ground is modeled by a quadratic equation. The initial velocity, $v_0$ , propels the object up until gravity causes the object to fall back down.

Projectile motion. The height in feet, $h$ , of an object shot upwards into the air with initial velocity, $v_0$ , after $t$ seconds is given by

h=-16t^2+v_0t.

We can use this formula to find how many seconds it will take for a firework to reach a specific height.

Example 9.39

A firework is shot upwards with initial velocity 130 feet per second. How many seconds will it take to reach a height of 260 feet? Round to the nearest tenth of a second.

Solution.

Step 1. Read the problem.

Step 2. Identify what we are looking for. We are looking for the number of seconds, which is time.

Step 3. Name what we are looking for. Let $t$ be the number of seconds.

Step 4. Translate into an equation. Use the projectile-motion formula.

Step 5. Solve the equation. Substitute $v_0=130$ and $h=260$ :

260=-16t^2+130t.

Rewrite it as $16t^2-130t+260=0$ , where $a=16$ , $b=-130$ , and $c=260$ .

\begin{aligned} t&=\frac{-(-130)\pm\sqrt{(-130)^2-4(16)(260)}}{2(16)}\\ &=\frac{130\pm\sqrt{260}}{32}. \end{aligned}

Thus $t\approx4.6$ seconds or $t\approx3.6$ seconds.

Step 6. Check the answer. The check is left to you.

Step 7. Answer the question. As the firework goes up, it will reach 260 feet after approximately 3.6 seconds. It will also pass that height on the way down at 4.6 seconds.

An arrow is shot from the ground at an initial speed of 108 ft/s. Use $h=-16t^2+v_0t$ to determine when it will be 180 feet from the ground. Enter both times to the nearest tenth, separated by a comma.

A man throws a ball into the air at 96 ft/s. Use $h=-16t^2+v_0t$ to determine when its height will be 48 feet. Enter both times to the nearest tenth, separated by a comma.

We have solved uniform motion problems using the formula $D=rt$ in previous chapters. We used a table to organize the information and lead us to the equation.

	Rate $\cdot$	Time $=$	Distance
	$r$	$t$	$D$

The formula $D=rt$ assumes we know $r$ and $t$ and use them to find $D$ . If we know $D$ and $r$ and need to find $t$ , we solve for $t$ and get $t=\tfrac Dr$ . Some uniform motion problems are also modeled by quadratic equations.

Example 9.40

Professor Smith just returned from a conference that was 2,000 miles east of his home. His total time in the airplane for the round trip was 9 hours. If the plane was flying at a rate of 450 miles per hour, what was the speed of the jet stream?

Solution. This is a uniform motion situation. A diagram helps us visualize the 2,000-mile trip with the wind and the 2,000-mile return trip against the wind.

We are looking for the speed of the jet stream. Let $r$ be its speed. With the wind, the plane’s rate is $450+r$ ; against the wind, it is $450-r$ .

Type	Rate	Time	Distance
Headwind	$450-r$	$\tfrac{2000}{450-r}$	$2000$
Tailwind	$450+r$	$\tfrac{2000}{450+r}$	$2000$

The times add to 9:

\frac{2000}{450-r}+\frac{2000}{450+r}=9.

Multiply both sides by the LCD, $(450-r)(450+r)$ , and solve:

\begin{aligned} 2000(450+r)+2000(450-r)&=9(450-r)(450+r)\\ 2000(900)&=9(450^2-r^2)\\ 2000(100)&=450^2-r^2\\ 200000&=202500-r^2\\ -2500&=-r^2\\ r&=50. \end{aligned}

Check: the tailwind rate is $500$ mph, so the trip takes $\tfrac{2000}{500}=4$ hours. The headwind rate is $400$ mph, so the return takes $\tfrac{2000}{400}=5$ hours. The times add to 9 hours. The speed of the jet stream was 50 mph.

MaryAnne's destination is 2,400 miles from home and her total round-trip flight time was 10 hours. If the plane flew at 500 mph in still air, what was the speed of the jet stream?

Gerry's destination is 3,000 miles from home and his total round-trip flight time was 11 hours. If the plane flew at 550 mph in still air, what was the speed of the jet stream?

Work applications can also be modeled by quadratic equations. We will set them up using the same methods we used when we solved them with rational equations. We’ll use a similar scenario now.

Example 9.41

The weekly gossip magazine has a big story about the presidential election and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 12 hours more than Press #2 to do the job and when both presses are running they can print the job in 8 hours. How long does it take for each press to print the job alone?

Solution. This is a work problem. A chart will help us organize the information. Let $x$ be the number of hours for Press #2 to complete the job.

	Number of hours needed to complete the job	Part of job completed/hour
Press #1	$x+12$	$\tfrac1{x+12}$
Press #2	$x$	$\tfrac1x$
Together	$8$	$\tfrac18$

The part completed by Press #1 plus the part completed by Press #2 equals the amount completed together:

\frac1{x+12}+\frac1x=\frac18.

Multiply by the LCD, $8x(x+12)$ , and solve.

\begin{aligned} 8x+8(x+12)&=x(x+12)\\ 8x+8x+96&=x^2+12x\\ 0&=x^2-4x-96\\ 0&=(x-12)(x+8). \end{aligned}

So $x=12$ or $x=-8$ . Since negative hours do not make sense, use $x=12$ . Press #1 would take 24 hours and Press #2 would take 12 hours to do the job alone.

Press #1 takes 6 hours more than Press #2 to print a job. Together they print it in 4 hours. How long does each press take alone? Enter the times for Press #1 and Press #2, separated by a comma.

A red hose takes 3 hours more than a green hose to fill a hot tub. Together they fill it in 2 hours. How long does each hose take alone? Enter the red-hose and green-hose times, separated by a comma.

Key terms. No new key terms are introduced in this section.

Adapted from [*Intermediate Algebra 2e*, Section 9.5](https://openstax.org/books/intermediate-algebra-2e/pages/9-5-solve-applications-of-quadratic-equations) by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under [CC BY-NC-SA 4.0](https://creativecommons.org/licenses/by-nc-sa/4.0/). Access the original for free at [OpenStax](https://openstax.org/). Changes: adapted the source to interactive web format and converted Try It exercises to auto-graded questions.

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